Problem: Suppose we have a surface $S$ defined by the transformation $T$ for $\dfrac{-\pi}{2} < u < \dfrac{\pi}{2}$ and $-1 < v < 1$. $T(u, v) = (\cos(u), \sin(u), v)$ What is the surface area of $S$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi$ (Choice B) B $2\pi$ (Choice C) C $4\pi$ (Choice D) D $3\pi$
Explanation: Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{-1}^1 |T_u \times T_v| \, du \, dv$ In our case, we are finding the surface area of a half cylinder. $x$ $y$ $z$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = 1$ [Calculation] The final step is to evaluate the double integral. $\begin{aligned} A &= \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{-1}^1 |T_u \times T_v| \, du \, dv \\ \\ &= \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{-1}^1 \, du \, dv \\ \\ &= 2 \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \, dv \\ \\ &= 2\pi \end{aligned}$ In conclusion, the surface area of $S$ is $2\pi$.